Construct amplitude and phase line spectra for Prob. 19.4.

Question

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Answer 1

Textbook Question

Chapter 19, Problem 6P

Construct amplitude and phase line spectra for Prob. 19.4.

Expert Solution & Answer

To determine

To graph: The amplitude and phase line spectra for the sawtooth wave as shown in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 1

Explanation of Solution

Given Information: The sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 2

Formula used:

Consider $f(t)$ is a periodic function with period $T$ defined in the interval $0≤x≤T$

then the Fourier series expansion of the function,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)]$ with $ω0=2πT$ .

And the coefficients are defined by,

$a0=1T∫0Tf(t)dt ; ak=2T∫0Tf(t)cos(kω0t)dt ; bk=2T∫0Tf(t)sin(kω0t)dt$

Alternatively, the Fourier series can also be written as, $f(t)=a0+∑k=1∞[ckcos(kω0t−ϕk)]$

Here, the amplitude $ck$ and phase $ϕk$ for each term is defined as,

$ck2=ak2+bk2,ϕk=tan−1(bkak)$

Plot $ck$ and $ϕk$ in the frequency domain to plot the amplitude and phase line spectra.

Graph:

Consider the sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 3

Therefore, the sawtooth wave is a periodic function $f(t)$ with period $T$ in the interval $0≤t≤T$ and from the graph,

$f(t)$ is a straight line joining the points $(0,0)$ and $(T2,−1)$ in $0≤t≤T2$ .

$f(t)$ is a straight line joining the points $(T2,1)$ and $(T,0)$ in $T2≤t≤T$ .

Therefore, the sawtooth wave,

$f(t)={−2Tt0≤t≤T22−2TtT2≤t≤T$

Therefore, the Fourier series expansion of this function is,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)];ω0=2πT$

In the above expression, the coefficients are defined by,

Now, find $ak$ .

$ak=2T∫0Tf(t)cos(2kπTt)dt=2T∫0T2(−2Tt)cos(2kπTt)dt+2T∫T2T(2−2Tt)cos(2kπTt)dt=−4T2∫0T2tcos(2kπTt)dt+4T∫T2Tcos(2kπTt)dt−4T2∫T2Ttcos(2kπTt)dt$

Consider,

$I=∫tcos(2kπTt)dt=t∫cos(2kπTt)dt−∫{ddt(t)∫cos(2kπTt)dt}dt=tT2kπsin(2kπTt)−∫T2kπsin(2kπTt)dt=Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)$

Hence,

$∫0T2tcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]0T2=T2kπT2sin(2kπTT2)+T24k2π2cos(2kπTT2)−T24k2π2cos(0)=T24kπsin(kπ)+T24k2π2cos(kπ)−T24k2π2=T24k2π2(−1)k−T24k2π2$

Further,

$∫T2Tcos(2kπTt)dt=T2kπsin(2kπTt)|T2T=T2kπ{sin(2kπ)−sin(kπ)}=0$

$∫T2Ttcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]T2T=T22kπsin(2kπ)+T24k2π2cos(2kπ)−T24kπsin(kπ)−T24k2π2cos(kπ)=T24k2π2−T24k2π2(−1)k$

Therefore,

$ak=−4T2[T24k2π2(−1)k−T24k2π2]+4T×0−4T2[T24k2π2−T24k2π2(−1)k]=−1k2π2(−1)k+1k2π2−1k2π2+1k2π2(−1)k=0$

Now, find $bk$ .

$bk=2T∫0Tf(t)sin(2kπTt)dt=2T∫0T2(−2Tt)sin(2kπTt)dt+2T∫T2T(2−2Tt)sin(2kπTt)dt=−4T2∫0T2tsin(2kπTt)dt+4T∫T2Tsin(2kπTt)dt−4T2∫T2Ttsin(2kπTt)dt$

Consider,

$I=∫tsin(2kπTt)dt=t∫sin(2kπTt)dt−∫{ddt(t)∫sin(2kπTt)dt}dt=−tT2kπcos(2kπTt)+∫T2kπcos(2kπTt)dt=−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)$

Hence,

$∫0T2tsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]0T2=−T24kπcos(kπ)+T24k2π2sin(kπ)=−T24kπ(−1)k$

$∫T2Tsin(2kπTt)dt=−T2kπcos(2kπTt)|T2T=−T2kπcos(2kπ)+T2kπcos(kπ)=−T2kπ+T2kπ(−1)k$

Further,

$∫T2Ttsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]T2T=−T22kπcos(2kπ)+T24k2π2sin(2kπ)+T24kπcos(kπ)−T24k2π2sin(kπ)=−T22kπ+T24kπ(−1)k$

Thus,

$bk=−4T2[−T24kπ(−1)k]+4T[−T2kπ+T2kπ(−1)k]−4T2[−T22kπ+T24kπ(−1)k]=1kπ(−1)k−2kπ+2kπ(−1)k+2kπ−1kπ(−1)k=2kπ(−1)k$

Hence, the coefficients of the Fourier series expansions are,

$a0=0,ak=0,bk=2kπ(−1)k$

That is,

$bk={2kπ>0k even−2kπ<0k odd$

Consider,

$ck=ak2+bk2=0+4k2π2(−1)2k=4k2π2=2kπ$

Thus, the amplitude of the $kth$ term is $2kπ$ , that is, the amplitudes for $k=1,2,3,4,5,...$ are,

$2π,1π,23π,12π,25π,...$

Furthermore, consider,

$ϕk=tan−1(bkak)$

As $ak=0$ and $bk>0$ for even $k$ thus,

$ϕk=π2$

As $ak=0$ and $bk<0$ for odd $k$ thus,

$ϕk=−π2$

Therefore,

$ϕk={π2k even−π2k odd$

Thus, the phases corresponding to $k=1,2,3,4,5,...$ are $−π2,π2,−π2,π2,−π2,...$

Use the following MATLAB code to construct the amplitude plot.

function Code_97924_19_6P_a()

for k = 1: 8

f(k) = k;

% define the amplitude

A(k) = 2/(k*pi);

end

% plot the values thus obtained

stem(f,A,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[A(8) A(4) A(2) A(1)]);

set(gca,'YTickLabel',{'1/4\pi' '1/2\pi' '1/\pi' '2/\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

end

Execute the above code to obtain the amplitude plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 4

Interpretation: The above plot shows the amplitude plot for the sawtooth wave as shown in the figure provided.

Use the following MATLAB file can be used to construct the phase plot.

function Code_97924_19_6P_b()

for k = 1: 8

f(k) = k;

% define the phase

P(k) = (-1)^k*(pi/2);

end

% plot the values thus obtained

stem(f,P,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[-pi -pi/2 0 pi/2 pi]);

set(gca,'YTickLabel',{'-\pi' '-\pi/2' '0' '\pi/2' '\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

ylim([-pi-0.1, pi+0.1]); grid on

end

Execute the above code to obtain the plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 5

Interpretation: The above plot shows the phase line spectra for the sawtooth wave as shown in the figure provided.

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Answer 2

Textbook Question

Chapter 19, Problem 6P

Construct amplitude and phase line spectra for Prob. 19.4.

Expert Solution & Answer

To determine

To graph: The amplitude and phase line spectra for the sawtooth wave as shown in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 1

Explanation of Solution

Given Information: The sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 2

Formula used:

Consider $f(t)$ is a periodic function with period $T$ defined in the interval $0≤x≤T$

then the Fourier series expansion of the function,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)]$ with $ω0=2πT$ .

And the coefficients are defined by,

$a0=1T∫0Tf(t)dt ; ak=2T∫0Tf(t)cos(kω0t)dt ; bk=2T∫0Tf(t)sin(kω0t)dt$

Alternatively, the Fourier series can also be written as, $f(t)=a0+∑k=1∞[ckcos(kω0t−ϕk)]$

Here, the amplitude $ck$ and phase $ϕk$ for each term is defined as,

$ck2=ak2+bk2,ϕk=tan−1(bkak)$

Plot $ck$ and $ϕk$ in the frequency domain to plot the amplitude and phase line spectra.

Graph:

Consider the sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 3

Therefore, the sawtooth wave is a periodic function $f(t)$ with period $T$ in the interval $0≤t≤T$ and from the graph,

$f(t)$ is a straight line joining the points $(0,0)$ and $(T2,−1)$ in $0≤t≤T2$ .

$f(t)$ is a straight line joining the points $(T2,1)$ and $(T,0)$ in $T2≤t≤T$ .

Therefore, the sawtooth wave,

$f(t)={−2Tt0≤t≤T22−2TtT2≤t≤T$

Therefore, the Fourier series expansion of this function is,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)];ω0=2πT$

In the above expression, the coefficients are defined by,

Now, find $ak$ .

$ak=2T∫0Tf(t)cos(2kπTt)dt=2T∫0T2(−2Tt)cos(2kπTt)dt+2T∫T2T(2−2Tt)cos(2kπTt)dt=−4T2∫0T2tcos(2kπTt)dt+4T∫T2Tcos(2kπTt)dt−4T2∫T2Ttcos(2kπTt)dt$

Consider,

$I=∫tcos(2kπTt)dt=t∫cos(2kπTt)dt−∫{ddt(t)∫cos(2kπTt)dt}dt=tT2kπsin(2kπTt)−∫T2kπsin(2kπTt)dt=Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)$

Hence,

$∫0T2tcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]0T2=T2kπT2sin(2kπTT2)+T24k2π2cos(2kπTT2)−T24k2π2cos(0)=T24kπsin(kπ)+T24k2π2cos(kπ)−T24k2π2=T24k2π2(−1)k−T24k2π2$

Further,

$∫T2Tcos(2kπTt)dt=T2kπsin(2kπTt)|T2T=T2kπ{sin(2kπ)−sin(kπ)}=0$

$∫T2Ttcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]T2T=T22kπsin(2kπ)+T24k2π2cos(2kπ)−T24kπsin(kπ)−T24k2π2cos(kπ)=T24k2π2−T24k2π2(−1)k$

Therefore,

$ak=−4T2[T24k2π2(−1)k−T24k2π2]+4T×0−4T2[T24k2π2−T24k2π2(−1)k]=−1k2π2(−1)k+1k2π2−1k2π2+1k2π2(−1)k=0$

Now, find $bk$ .

$bk=2T∫0Tf(t)sin(2kπTt)dt=2T∫0T2(−2Tt)sin(2kπTt)dt+2T∫T2T(2−2Tt)sin(2kπTt)dt=−4T2∫0T2tsin(2kπTt)dt+4T∫T2Tsin(2kπTt)dt−4T2∫T2Ttsin(2kπTt)dt$

Consider,

$I=∫tsin(2kπTt)dt=t∫sin(2kπTt)dt−∫{ddt(t)∫sin(2kπTt)dt}dt=−tT2kπcos(2kπTt)+∫T2kπcos(2kπTt)dt=−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)$

Hence,

$∫0T2tsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]0T2=−T24kπcos(kπ)+T24k2π2sin(kπ)=−T24kπ(−1)k$

$∫T2Tsin(2kπTt)dt=−T2kπcos(2kπTt)|T2T=−T2kπcos(2kπ)+T2kπcos(kπ)=−T2kπ+T2kπ(−1)k$

Further,

$∫T2Ttsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]T2T=−T22kπcos(2kπ)+T24k2π2sin(2kπ)+T24kπcos(kπ)−T24k2π2sin(kπ)=−T22kπ+T24kπ(−1)k$

Thus,

$bk=−4T2[−T24kπ(−1)k]+4T[−T2kπ+T2kπ(−1)k]−4T2[−T22kπ+T24kπ(−1)k]=1kπ(−1)k−2kπ+2kπ(−1)k+2kπ−1kπ(−1)k=2kπ(−1)k$

Hence, the coefficients of the Fourier series expansions are,

$a0=0,ak=0,bk=2kπ(−1)k$

That is,

$bk={2kπ>0k even−2kπ<0k odd$

Consider,

$ck=ak2+bk2=0+4k2π2(−1)2k=4k2π2=2kπ$

Thus, the amplitude of the $kth$ term is $2kπ$ , that is, the amplitudes for $k=1,2,3,4,5,...$ are,

$2π,1π,23π,12π,25π,...$

Furthermore, consider,

$ϕk=tan−1(bkak)$

As $ak=0$ and $bk>0$ for even $k$ thus,

$ϕk=π2$

As $ak=0$ and $bk<0$ for odd $k$ thus,

$ϕk=−π2$

Therefore,

$ϕk={π2k even−π2k odd$

Thus, the phases corresponding to $k=1,2,3,4,5,...$ are $−π2,π2,−π2,π2,−π2,...$

Use the following MATLAB code to construct the amplitude plot.

function Code_97924_19_6P_a()

for k = 1: 8

f(k) = k;

% define the amplitude

A(k) = 2/(k*pi);

end

% plot the values thus obtained

stem(f,A,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[A(8) A(4) A(2) A(1)]);

set(gca,'YTickLabel',{'1/4\pi' '1/2\pi' '1/\pi' '2/\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

end

Execute the above code to obtain the amplitude plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 4

Interpretation: The above plot shows the amplitude plot for the sawtooth wave as shown in the figure provided.

Use the following MATLAB file can be used to construct the phase plot.

function Code_97924_19_6P_b()

for k = 1: 8

f(k) = k;

% define the phase

P(k) = (-1)^k*(pi/2);

end

% plot the values thus obtained

stem(f,P,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[-pi -pi/2 0 pi/2 pi]);

set(gca,'YTickLabel',{'-\pi' '-\pi/2' '0' '\pi/2' '\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

ylim([-pi-0.1, pi+0.1]); grid on

end

Execute the above code to obtain the plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 5

Interpretation: The above plot shows the phase line spectra for the sawtooth wave as shown in the figure provided.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 19, Problem 6P

Construct amplitude and phase line spectra for Prob. 19.4.

Expert Solution & Answer

To determine

To graph: The amplitude and phase line spectra for the sawtooth wave as shown in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 1

Explanation of Solution

Given Information: The sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 2

Formula used:

Consider $f(t)$ is a periodic function with period $T$ defined in the interval $0≤x≤T$

then the Fourier series expansion of the function,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)]$ with $ω0=2πT$ .

And the coefficients are defined by,

$a0=1T∫0Tf(t)dt ; ak=2T∫0Tf(t)cos(kω0t)dt ; bk=2T∫0Tf(t)sin(kω0t)dt$

Alternatively, the Fourier series can also be written as, $f(t)=a0+∑k=1∞[ckcos(kω0t−ϕk)]$

Here, the amplitude $ck$ and phase $ϕk$ for each term is defined as,

$ck2=ak2+bk2,ϕk=tan−1(bkak)$

Plot $ck$ and $ϕk$ in the frequency domain to plot the amplitude and phase line spectra.

Graph:

Consider the sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 3

Therefore, the sawtooth wave is a periodic function $f(t)$ with period $T$ in the interval $0≤t≤T$ and from the graph,

$f(t)$ is a straight line joining the points $(0,0)$ and $(T2,−1)$ in $0≤t≤T2$ .

$f(t)$ is a straight line joining the points $(T2,1)$ and $(T,0)$ in $T2≤t≤T$ .

Therefore, the sawtooth wave,

$f(t)={−2Tt0≤t≤T22−2TtT2≤t≤T$

Therefore, the Fourier series expansion of this function is,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)];ω0=2πT$

In the above expression, the coefficients are defined by,

Now, find $ak$ .

$ak=2T∫0Tf(t)cos(2kπTt)dt=2T∫0T2(−2Tt)cos(2kπTt)dt+2T∫T2T(2−2Tt)cos(2kπTt)dt=−4T2∫0T2tcos(2kπTt)dt+4T∫T2Tcos(2kπTt)dt−4T2∫T2Ttcos(2kπTt)dt$

Consider,

$I=∫tcos(2kπTt)dt=t∫cos(2kπTt)dt−∫{ddt(t)∫cos(2kπTt)dt}dt=tT2kπsin(2kπTt)−∫T2kπsin(2kπTt)dt=Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)$

Hence,

$∫0T2tcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]0T2=T2kπT2sin(2kπTT2)+T24k2π2cos(2kπTT2)−T24k2π2cos(0)=T24kπsin(kπ)+T24k2π2cos(kπ)−T24k2π2=T24k2π2(−1)k−T24k2π2$

Further,

$∫T2Tcos(2kπTt)dt=T2kπsin(2kπTt)|T2T=T2kπ{sin(2kπ)−sin(kπ)}=0$

$∫T2Ttcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]T2T=T22kπsin(2kπ)+T24k2π2cos(2kπ)−T24kπsin(kπ)−T24k2π2cos(kπ)=T24k2π2−T24k2π2(−1)k$

Therefore,

$ak=−4T2[T24k2π2(−1)k−T24k2π2]+4T×0−4T2[T24k2π2−T24k2π2(−1)k]=−1k2π2(−1)k+1k2π2−1k2π2+1k2π2(−1)k=0$

Now, find $bk$ .

$bk=2T∫0Tf(t)sin(2kπTt)dt=2T∫0T2(−2Tt)sin(2kπTt)dt+2T∫T2T(2−2Tt)sin(2kπTt)dt=−4T2∫0T2tsin(2kπTt)dt+4T∫T2Tsin(2kπTt)dt−4T2∫T2Ttsin(2kπTt)dt$

Consider,

$I=∫tsin(2kπTt)dt=t∫sin(2kπTt)dt−∫{ddt(t)∫sin(2kπTt)dt}dt=−tT2kπcos(2kπTt)+∫T2kπcos(2kπTt)dt=−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)$

Hence,

$∫0T2tsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]0T2=−T24kπcos(kπ)+T24k2π2sin(kπ)=−T24kπ(−1)k$

$∫T2Tsin(2kπTt)dt=−T2kπcos(2kπTt)|T2T=−T2kπcos(2kπ)+T2kπcos(kπ)=−T2kπ+T2kπ(−1)k$

Further,

$∫T2Ttsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]T2T=−T22kπcos(2kπ)+T24k2π2sin(2kπ)+T24kπcos(kπ)−T24k2π2sin(kπ)=−T22kπ+T24kπ(−1)k$

Thus,

$bk=−4T2[−T24kπ(−1)k]+4T[−T2kπ+T2kπ(−1)k]−4T2[−T22kπ+T24kπ(−1)k]=1kπ(−1)k−2kπ+2kπ(−1)k+2kπ−1kπ(−1)k=2kπ(−1)k$

Hence, the coefficients of the Fourier series expansions are,

$a0=0,ak=0,bk=2kπ(−1)k$

That is,

$bk={2kπ>0k even−2kπ<0k odd$

Consider,

$ck=ak2+bk2=0+4k2π2(−1)2k=4k2π2=2kπ$

Thus, the amplitude of the $kth$ term is $2kπ$ , that is, the amplitudes for $k=1,2,3,4,5,...$ are,

$2π,1π,23π,12π,25π,...$

Furthermore, consider,

$ϕk=tan−1(bkak)$

As $ak=0$ and $bk>0$ for even $k$ thus,

$ϕk=π2$

As $ak=0$ and $bk<0$ for odd $k$ thus,

$ϕk=−π2$

Therefore,

$ϕk={π2k even−π2k odd$

Thus, the phases corresponding to $k=1,2,3,4,5,...$ are $−π2,π2,−π2,π2,−π2,...$

Use the following MATLAB code to construct the amplitude plot.

function Code_97924_19_6P_a()

for k = 1: 8

f(k) = k;

% define the amplitude

A(k) = 2/(k*pi);

end

% plot the values thus obtained

stem(f,A,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[A(8) A(4) A(2) A(1)]);

set(gca,'YTickLabel',{'1/4\pi' '1/2\pi' '1/\pi' '2/\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

end

Execute the above code to obtain the amplitude plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 4

Interpretation: The above plot shows the amplitude plot for the sawtooth wave as shown in the figure provided.

Use the following MATLAB file can be used to construct the phase plot.

function Code_97924_19_6P_b()

for k = 1: 8

f(k) = k;

% define the phase

P(k) = (-1)^k*(pi/2);

end

% plot the values thus obtained

stem(f,P,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[-pi -pi/2 0 pi/2 pi]);

set(gca,'YTickLabel',{'-\pi' '-\pi/2' '0' '\pi/2' '\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

ylim([-pi-0.1, pi+0.1]); grid on

end

Execute the above code to obtain the plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 5

Interpretation: The above plot shows the phase line spectra for the sawtooth wave as shown in the figure provided.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 19, Problem 6P

Construct amplitude and phase line spectra for Prob. 19.4.

Expert Solution & Answer

To determine

To graph: The amplitude and phase line spectra for the sawtooth wave as shown in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 1

Explanation of Solution

Given Information: The sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 2

Formula used:

Consider $f(t)$ is a periodic function with period $T$ defined in the interval $0≤x≤T$

then the Fourier series expansion of the function,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)]$ with $ω0=2πT$ .

And the coefficients are defined by,

$a0=1T∫0Tf(t)dt ; ak=2T∫0Tf(t)cos(kω0t)dt ; bk=2T∫0Tf(t)sin(kω0t)dt$

Alternatively, the Fourier series can also be written as, $f(t)=a0+∑k=1∞[ckcos(kω0t−ϕk)]$

Here, the amplitude $ck$ and phase $ϕk$ for each term is defined as,

$ck2=ak2+bk2,ϕk=tan−1(bkak)$

Plot $ck$ and $ϕk$ in the frequency domain to plot the amplitude and phase line spectra.

Graph:

Consider the sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 3

Therefore, the sawtooth wave is a periodic function $f(t)$ with period $T$ in the interval $0≤t≤T$ and from the graph,

$f(t)$ is a straight line joining the points $(0,0)$ and $(T2,−1)$ in $0≤t≤T2$ .

$f(t)$ is a straight line joining the points $(T2,1)$ and $(T,0)$ in $T2≤t≤T$ .

Therefore, the sawtooth wave,

$f(t)={−2Tt0≤t≤T22−2TtT2≤t≤T$

Therefore, the Fourier series expansion of this function is,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)];ω0=2πT$

In the above expression, the coefficients are defined by,

Now, find $ak$ .

$ak=2T∫0Tf(t)cos(2kπTt)dt=2T∫0T2(−2Tt)cos(2kπTt)dt+2T∫T2T(2−2Tt)cos(2kπTt)dt=−4T2∫0T2tcos(2kπTt)dt+4T∫T2Tcos(2kπTt)dt−4T2∫T2Ttcos(2kπTt)dt$

Consider,

$I=∫tcos(2kπTt)dt=t∫cos(2kπTt)dt−∫{ddt(t)∫cos(2kπTt)dt}dt=tT2kπsin(2kπTt)−∫T2kπsin(2kπTt)dt=Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)$

Hence,

$∫0T2tcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]0T2=T2kπT2sin(2kπTT2)+T24k2π2cos(2kπTT2)−T24k2π2cos(0)=T24kπsin(kπ)+T24k2π2cos(kπ)−T24k2π2=T24k2π2(−1)k−T24k2π2$

Further,

$∫T2Tcos(2kπTt)dt=T2kπsin(2kπTt)|T2T=T2kπ{sin(2kπ)−sin(kπ)}=0$

$∫T2Ttcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]T2T=T22kπsin(2kπ)+T24k2π2cos(2kπ)−T24kπsin(kπ)−T24k2π2cos(kπ)=T24k2π2−T24k2π2(−1)k$

Therefore,

$ak=−4T2[T24k2π2(−1)k−T24k2π2]+4T×0−4T2[T24k2π2−T24k2π2(−1)k]=−1k2π2(−1)k+1k2π2−1k2π2+1k2π2(−1)k=0$

Now, find $bk$ .

$bk=2T∫0Tf(t)sin(2kπTt)dt=2T∫0T2(−2Tt)sin(2kπTt)dt+2T∫T2T(2−2Tt)sin(2kπTt)dt=−4T2∫0T2tsin(2kπTt)dt+4T∫T2Tsin(2kπTt)dt−4T2∫T2Ttsin(2kπTt)dt$

Consider,

$I=∫tsin(2kπTt)dt=t∫sin(2kπTt)dt−∫{ddt(t)∫sin(2kπTt)dt}dt=−tT2kπcos(2kπTt)+∫T2kπcos(2kπTt)dt=−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)$

Hence,

$∫0T2tsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]0T2=−T24kπcos(kπ)+T24k2π2sin(kπ)=−T24kπ(−1)k$

$∫T2Tsin(2kπTt)dt=−T2kπcos(2kπTt)|T2T=−T2kπcos(2kπ)+T2kπcos(kπ)=−T2kπ+T2kπ(−1)k$

Further,

$∫T2Ttsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]T2T=−T22kπcos(2kπ)+T24k2π2sin(2kπ)+T24kπcos(kπ)−T24k2π2sin(kπ)=−T22kπ+T24kπ(−1)k$

Thus,

$bk=−4T2[−T24kπ(−1)k]+4T[−T2kπ+T2kπ(−1)k]−4T2[−T22kπ+T24kπ(−1)k]=1kπ(−1)k−2kπ+2kπ(−1)k+2kπ−1kπ(−1)k=2kπ(−1)k$

Hence, the coefficients of the Fourier series expansions are,

$a0=0,ak=0,bk=2kπ(−1)k$

That is,

$bk={2kπ>0k even−2kπ<0k odd$

Consider,

$ck=ak2+bk2=0+4k2π2(−1)2k=4k2π2=2kπ$

Thus, the amplitude of the $kth$ term is $2kπ$ , that is, the amplitudes for $k=1,2,3,4,5,...$ are,

$2π,1π,23π,12π,25π,...$

Furthermore, consider,

$ϕk=tan−1(bkak)$

As $ak=0$ and $bk>0$ for even $k$ thus,

$ϕk=π2$

As $ak=0$ and $bk<0$ for odd $k$ thus,

$ϕk=−π2$

Therefore,

$ϕk={π2k even−π2k odd$

Thus, the phases corresponding to $k=1,2,3,4,5,...$ are $−π2,π2,−π2,π2,−π2,...$

Use the following MATLAB code to construct the amplitude plot.

function Code_97924_19_6P_a()

for k = 1: 8

f(k) = k;

% define the amplitude

A(k) = 2/(k*pi);

end

% plot the values thus obtained

stem(f,A,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[A(8) A(4) A(2) A(1)]);

set(gca,'YTickLabel',{'1/4\pi' '1/2\pi' '1/\pi' '2/\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

end

Execute the above code to obtain the amplitude plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 4

Interpretation: The above plot shows the amplitude plot for the sawtooth wave as shown in the figure provided.

Use the following MATLAB file can be used to construct the phase plot.

function Code_97924_19_6P_b()

for k = 1: 8

f(k) = k;

% define the phase

P(k) = (-1)^k*(pi/2);

end

% plot the values thus obtained

stem(f,P,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[-pi -pi/2 0 pi/2 pi]);

set(gca,'YTickLabel',{'-\pi' '-\pi/2' '0' '\pi/2' '\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

ylim([-pi-0.1, pi+0.1]); grid on

end

Execute the above code to obtain the plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 5

Interpretation: The above plot shows the phase line spectra for the sawtooth wave as shown in the figure provided.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To graph: The amplitude and phase line spectra for the sawtooth wave as shown in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 1

Explanation of Solution

Given Information: The sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 2

Formula used:

Consider $f(t)$ is a periodic function with period $T$ defined in the interval $0≤x≤T$

then the Fourier series expansion of the function,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)]$ with $ω0=2πT$ .

And the coefficients are defined by,

$a0=1T∫0Tf(t)dt ; ak=2T∫0Tf(t)cos(kω0t)dt ; bk=2T∫0Tf(t)sin(kω0t)dt$

Alternatively, the Fourier series can also be written as, $f(t)=a0+∑k=1∞[ckcos(kω0t−ϕk)]$

Here, the amplitude $ck$ and phase $ϕk$ for each term is defined as,

$ck2=ak2+bk2,ϕk=tan−1(bkak)$

Plot $ck$ and $ϕk$ in the frequency domain to plot the amplitude and phase line spectra.

Graph:

Consider the sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 3

Therefore, the sawtooth wave is a periodic function $f(t)$ with period $T$ in the interval $0≤t≤T$ and from the graph,

$f(t)$ is a straight line joining the points $(0,0)$ and $(T2,−1)$ in $0≤t≤T2$ .

$f(t)$ is a straight line joining the points $(T2,1)$ and $(T,0)$ in $T2≤t≤T$ .

Therefore, the sawtooth wave,

$f(t)={−2Tt0≤t≤T22−2TtT2≤t≤T$

Therefore, the Fourier series expansion of this function is,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)];ω0=2πT$

In the above expression, the coefficients are defined by,

Now, find $ak$ .

$ak=2T∫0Tf(t)cos(2kπTt)dt=2T∫0T2(−2Tt)cos(2kπTt)dt+2T∫T2T(2−2Tt)cos(2kπTt)dt=−4T2∫0T2tcos(2kπTt)dt+4T∫T2Tcos(2kπTt)dt−4T2∫T2Ttcos(2kπTt)dt$

Consider,

$I=∫tcos(2kπTt)dt=t∫cos(2kπTt)dt−∫{ddt(t)∫cos(2kπTt)dt}dt=tT2kπsin(2kπTt)−∫T2kπsin(2kπTt)dt=Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)$

Hence,

$∫0T2tcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]0T2=T2kπT2sin(2kπTT2)+T24k2π2cos(2kπTT2)−T24k2π2cos(0)=T24kπsin(kπ)+T24k2π2cos(kπ)−T24k2π2=T24k2π2(−1)k−T24k2π2$

Further,

$∫T2Tcos(2kπTt)dt=T2kπsin(2kπTt)|T2T=T2kπ{sin(2kπ)−sin(kπ)}=0$

$∫T2Ttcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]T2T=T22kπsin(2kπ)+T24k2π2cos(2kπ)−T24kπsin(kπ)−T24k2π2cos(kπ)=T24k2π2−T24k2π2(−1)k$

Therefore,

$ak=−4T2[T24k2π2(−1)k−T24k2π2]+4T×0−4T2[T24k2π2−T24k2π2(−1)k]=−1k2π2(−1)k+1k2π2−1k2π2+1k2π2(−1)k=0$

Now, find $bk$ .

$bk=2T∫0Tf(t)sin(2kπTt)dt=2T∫0T2(−2Tt)sin(2kπTt)dt+2T∫T2T(2−2Tt)sin(2kπTt)dt=−4T2∫0T2tsin(2kπTt)dt+4T∫T2Tsin(2kπTt)dt−4T2∫T2Ttsin(2kπTt)dt$

Consider,

$I=∫tsin(2kπTt)dt=t∫sin(2kπTt)dt−∫{ddt(t)∫sin(2kπTt)dt}dt=−tT2kπcos(2kπTt)+∫T2kπcos(2kπTt)dt=−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)$

Hence,

$∫0T2tsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]0T2=−T24kπcos(kπ)+T24k2π2sin(kπ)=−T24kπ(−1)k$

$∫T2Tsin(2kπTt)dt=−T2kπcos(2kπTt)|T2T=−T2kπcos(2kπ)+T2kπcos(kπ)=−T2kπ+T2kπ(−1)k$

Further,

$∫T2Ttsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]T2T=−T22kπcos(2kπ)+T24k2π2sin(2kπ)+T24kπcos(kπ)−T24k2π2sin(kπ)=−T22kπ+T24kπ(−1)k$

Thus,

$bk=−4T2[−T24kπ(−1)k]+4T[−T2kπ+T2kπ(−1)k]−4T2[−T22kπ+T24kπ(−1)k]=1kπ(−1)k−2kπ+2kπ(−1)k+2kπ−1kπ(−1)k=2kπ(−1)k$

Hence, the coefficients of the Fourier series expansions are,

$a0=0,ak=0,bk=2kπ(−1)k$

That is,

$bk={2kπ>0k even−2kπ<0k odd$

Consider,

$ck=ak2+bk2=0+4k2π2(−1)2k=4k2π2=2kπ$

Thus, the amplitude of the $kth$ term is $2kπ$ , that is, the amplitudes for $k=1,2,3,4,5,...$ are,

$2π,1π,23π,12π,25π,...$

Furthermore, consider,

$ϕk=tan−1(bkak)$

As $ak=0$ and $bk>0$ for even $k$ thus,

$ϕk=π2$

As $ak=0$ and $bk<0$ for odd $k$ thus,

$ϕk=−π2$

Therefore,

$ϕk={π2k even−π2k odd$

Thus, the phases corresponding to $k=1,2,3,4,5,...$ are $−π2,π2,−π2,π2,−π2,...$

Use the following MATLAB code to construct the amplitude plot.

function Code_97924_19_6P_a()

for k = 1: 8

f(k) = k;

% define the amplitude

A(k) = 2/(k*pi);

end

% plot the values thus obtained

stem(f,A,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[A(8) A(4) A(2) A(1)]);

set(gca,'YTickLabel',{'1/4\pi' '1/2\pi' '1/\pi' '2/\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

end

Execute the above code to obtain the amplitude plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 4

Interpretation: The above plot shows the amplitude plot for the sawtooth wave as shown in the figure provided.

Use the following MATLAB file can be used to construct the phase plot.

function Code_97924_19_6P_b()

for k = 1: 8

f(k) = k;

% define the phase

P(k) = (-1)^k*(pi/2);

end

% plot the values thus obtained

stem(f,P,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[-pi -pi/2 0 pi/2 pi]);

set(gca,'YTickLabel',{'-\pi' '-\pi/2' '0' '\pi/2' '\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

ylim([-pi-0.1, pi+0.1]); grid on

end

Execute the above code to obtain the plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 5

Interpretation: The above plot shows the phase line spectra for the sawtooth wave as shown in the figure provided.

Answer 8

Expert Solution & Answer

To determine

To graph: The amplitude and phase line spectra for the sawtooth wave as shown in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 1

Explanation of Solution

Given Information: The sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 2

Formula used:

Consider $f(t)$ is a periodic function with period $T$ defined in the interval $0≤x≤T$

then the Fourier series expansion of the function,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)]$ with $ω0=2πT$ .

And the coefficients are defined by,

$a0=1T∫0Tf(t)dt ; ak=2T∫0Tf(t)cos(kω0t)dt ; bk=2T∫0Tf(t)sin(kω0t)dt$

Alternatively, the Fourier series can also be written as, $f(t)=a0+∑k=1∞[ckcos(kω0t−ϕk)]$

Here, the amplitude $ck$ and phase $ϕk$ for each term is defined as,

$ck2=ak2+bk2,ϕk=tan−1(bkak)$

Plot $ck$ and $ϕk$ in the frequency domain to plot the amplitude and phase line spectra.

Graph:

Consider the sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 3

Therefore, the sawtooth wave is a periodic function $f(t)$ with period $T$ in the interval $0≤t≤T$ and from the graph,

$f(t)$ is a straight line joining the points $(0,0)$ and $(T2,−1)$ in $0≤t≤T2$ .

$f(t)$ is a straight line joining the points $(T2,1)$ and $(T,0)$ in $T2≤t≤T$ .

Therefore, the sawtooth wave,

$f(t)={−2Tt0≤t≤T22−2TtT2≤t≤T$

Therefore, the Fourier series expansion of this function is,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)];ω0=2πT$

In the above expression, the coefficients are defined by,

Now, find $ak$ .

$ak=2T∫0Tf(t)cos(2kπTt)dt=2T∫0T2(−2Tt)cos(2kπTt)dt+2T∫T2T(2−2Tt)cos(2kπTt)dt=−4T2∫0T2tcos(2kπTt)dt+4T∫T2Tcos(2kπTt)dt−4T2∫T2Ttcos(2kπTt)dt$

Consider,

$I=∫tcos(2kπTt)dt=t∫cos(2kπTt)dt−∫{ddt(t)∫cos(2kπTt)dt}dt=tT2kπsin(2kπTt)−∫T2kπsin(2kπTt)dt=Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)$

Hence,

$∫0T2tcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]0T2=T2kπT2sin(2kπTT2)+T24k2π2cos(2kπTT2)−T24k2π2cos(0)=T24kπsin(kπ)+T24k2π2cos(kπ)−T24k2π2=T24k2π2(−1)k−T24k2π2$

Further,

$∫T2Tcos(2kπTt)dt=T2kπsin(2kπTt)|T2T=T2kπ{sin(2kπ)−sin(kπ)}=0$

$∫T2Ttcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]T2T=T22kπsin(2kπ)+T24k2π2cos(2kπ)−T24kπsin(kπ)−T24k2π2cos(kπ)=T24k2π2−T24k2π2(−1)k$

Therefore,

$ak=−4T2[T24k2π2(−1)k−T24k2π2]+4T×0−4T2[T24k2π2−T24k2π2(−1)k]=−1k2π2(−1)k+1k2π2−1k2π2+1k2π2(−1)k=0$

Now, find $bk$ .

$bk=2T∫0Tf(t)sin(2kπTt)dt=2T∫0T2(−2Tt)sin(2kπTt)dt+2T∫T2T(2−2Tt)sin(2kπTt)dt=−4T2∫0T2tsin(2kπTt)dt+4T∫T2Tsin(2kπTt)dt−4T2∫T2Ttsin(2kπTt)dt$

Consider,

$I=∫tsin(2kπTt)dt=t∫sin(2kπTt)dt−∫{ddt(t)∫sin(2kπTt)dt}dt=−tT2kπcos(2kπTt)+∫T2kπcos(2kπTt)dt=−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)$

Hence,

$∫0T2tsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]0T2=−T24kπcos(kπ)+T24k2π2sin(kπ)=−T24kπ(−1)k$

$∫T2Tsin(2kπTt)dt=−T2kπcos(2kπTt)|T2T=−T2kπcos(2kπ)+T2kπcos(kπ)=−T2kπ+T2kπ(−1)k$

Further,

$∫T2Ttsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]T2T=−T22kπcos(2kπ)+T24k2π2sin(2kπ)+T24kπcos(kπ)−T24k2π2sin(kπ)=−T22kπ+T24kπ(−1)k$

Thus,

$bk=−4T2[−T24kπ(−1)k]+4T[−T2kπ+T2kπ(−1)k]−4T2[−T22kπ+T24kπ(−1)k]=1kπ(−1)k−2kπ+2kπ(−1)k+2kπ−1kπ(−1)k=2kπ(−1)k$

Hence, the coefficients of the Fourier series expansions are,

$a0=0,ak=0,bk=2kπ(−1)k$

That is,

$bk={2kπ>0k even−2kπ<0k odd$

Consider,

$ck=ak2+bk2=0+4k2π2(−1)2k=4k2π2=2kπ$

Thus, the amplitude of the $kth$ term is $2kπ$ , that is, the amplitudes for $k=1,2,3,4,5,...$ are,

$2π,1π,23π,12π,25π,...$

Furthermore, consider,

$ϕk=tan−1(bkak)$

As $ak=0$ and $bk>0$ for even $k$ thus,

$ϕk=π2$

As $ak=0$ and $bk<0$ for odd $k$ thus,

$ϕk=−π2$

Therefore,

$ϕk={π2k even−π2k odd$

Thus, the phases corresponding to $k=1,2,3,4,5,...$ are $−π2,π2,−π2,π2,−π2,...$

Use the following MATLAB code to construct the amplitude plot.

function Code_97924_19_6P_a()

for k = 1: 8

f(k) = k;

% define the amplitude

A(k) = 2/(k*pi);

end

% plot the values thus obtained

stem(f,A,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[A(8) A(4) A(2) A(1)]);

set(gca,'YTickLabel',{'1/4\pi' '1/2\pi' '1/\pi' '2/\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

end

Execute the above code to obtain the amplitude plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 4

Interpretation: The above plot shows the amplitude plot for the sawtooth wave as shown in the figure provided.

Use the following MATLAB file can be used to construct the phase plot.

function Code_97924_19_6P_b()

for k = 1: 8

f(k) = k;

% define the phase

P(k) = (-1)^k*(pi/2);

end

% plot the values thus obtained

stem(f,P,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[-pi -pi/2 0 pi/2 pi]);

set(gca,'YTickLabel',{'-\pi' '-\pi/2' '0' '\pi/2' '\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

ylim([-pi-0.1, pi+0.1]); grid on

end

Execute the above code to obtain the plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 5

Interpretation: The above plot shows the phase line spectra for the sawtooth wave as shown in the figure provided.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To graph: The amplitude and phase line spectra for the sawtooth wave as shown in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 1

Answer 12

Explanation of Solution

Given Information: The sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 2

Formula used:

Consider $f(t)$ is a periodic function with period $T$ defined in the interval $0≤x≤T$

then the Fourier series expansion of the function,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)]$ with $ω0=2πT$ .

And the coefficients are defined by,

$a0=1T∫0Tf(t)dt ; ak=2T∫0Tf(t)cos(kω0t)dt ; bk=2T∫0Tf(t)sin(kω0t)dt$

Alternatively, the Fourier series can also be written as, $f(t)=a0+∑k=1∞[ckcos(kω0t−ϕk)]$

Here, the amplitude $ck$ and phase $ϕk$ for each term is defined as,

$ck2=ak2+bk2,ϕk=tan−1(bkak)$

Plot $ck$ and $ϕk$ in the frequency domain to plot the amplitude and phase line spectra.

Graph:

Consider the sawtooth wave given in the following figure,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 3

Therefore, the sawtooth wave is a periodic function $f(t)$ with period $T$ in the interval $0≤t≤T$ and from the graph,

$f(t)$ is a straight line joining the points $(0,0)$ and $(T2,−1)$ in $0≤t≤T2$ .

$f(t)$ is a straight line joining the points $(T2,1)$ and $(T,0)$ in $T2≤t≤T$ .

Therefore, the sawtooth wave,

$f(t)={−2Tt0≤t≤T22−2TtT2≤t≤T$

Therefore, the Fourier series expansion of this function is,

$f(t)=a0+∑k=1∞[akcos(kω0t)+bksin(kω0t)];ω0=2πT$

In the above expression, the coefficients are defined by,

Now, find $ak$ .

$ak=2T∫0Tf(t)cos(2kπTt)dt=2T∫0T2(−2Tt)cos(2kπTt)dt+2T∫T2T(2−2Tt)cos(2kπTt)dt=−4T2∫0T2tcos(2kπTt)dt+4T∫T2Tcos(2kπTt)dt−4T2∫T2Ttcos(2kπTt)dt$

Consider,

$I=∫tcos(2kπTt)dt=t∫cos(2kπTt)dt−∫{ddt(t)∫cos(2kπTt)dt}dt=tT2kπsin(2kπTt)−∫T2kπsin(2kπTt)dt=Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)$

Hence,

$∫0T2tcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]0T2=T2kπT2sin(2kπTT2)+T24k2π2cos(2kπTT2)−T24k2π2cos(0)=T24kπsin(kπ)+T24k2π2cos(kπ)−T24k2π2=T24k2π2(−1)k−T24k2π2$

Further,

$∫T2Tcos(2kπTt)dt=T2kπsin(2kπTt)|T2T=T2kπ{sin(2kπ)−sin(kπ)}=0$

$∫T2Ttcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]T2T=T22kπsin(2kπ)+T24k2π2cos(2kπ)−T24kπsin(kπ)−T24k2π2cos(kπ)=T24k2π2−T24k2π2(−1)k$

Therefore,

$ak=−4T2[T24k2π2(−1)k−T24k2π2]+4T×0−4T2[T24k2π2−T24k2π2(−1)k]=−1k2π2(−1)k+1k2π2−1k2π2+1k2π2(−1)k=0$

Now, find $bk$ .

$bk=2T∫0Tf(t)sin(2kπTt)dt=2T∫0T2(−2Tt)sin(2kπTt)dt+2T∫T2T(2−2Tt)sin(2kπTt)dt=−4T2∫0T2tsin(2kπTt)dt+4T∫T2Tsin(2kπTt)dt−4T2∫T2Ttsin(2kπTt)dt$

Consider,

$I=∫tsin(2kπTt)dt=t∫sin(2kπTt)dt−∫{ddt(t)∫sin(2kπTt)dt}dt=−tT2kπcos(2kπTt)+∫T2kπcos(2kπTt)dt=−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)$

Hence,

$∫0T2tsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]0T2=−T24kπcos(kπ)+T24k2π2sin(kπ)=−T24kπ(−1)k$

$∫T2Tsin(2kπTt)dt=−T2kπcos(2kπTt)|T2T=−T2kπcos(2kπ)+T2kπcos(kπ)=−T2kπ+T2kπ(−1)k$

Further,

$∫T2Ttsin(2kπTt)dt=[−Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]T2T=−T22kπcos(2kπ)+T24k2π2sin(2kπ)+T24kπcos(kπ)−T24k2π2sin(kπ)=−T22kπ+T24kπ(−1)k$

Thus,

$bk=−4T2[−T24kπ(−1)k]+4T[−T2kπ+T2kπ(−1)k]−4T2[−T22kπ+T24kπ(−1)k]=1kπ(−1)k−2kπ+2kπ(−1)k+2kπ−1kπ(−1)k=2kπ(−1)k$

Hence, the coefficients of the Fourier series expansions are,

$a0=0,ak=0,bk=2kπ(−1)k$

That is,

$bk={2kπ>0k even−2kπ<0k odd$

Consider,

$ck=ak2+bk2=0+4k2π2(−1)2k=4k2π2=2kπ$

Thus, the amplitude of the $kth$ term is $2kπ$ , that is, the amplitudes for $k=1,2,3,4,5,...$ are,

$2π,1π,23π,12π,25π,...$

Furthermore, consider,

$ϕk=tan−1(bkak)$

As $ak=0$ and $bk>0$ for even $k$ thus,

$ϕk=π2$

As $ak=0$ and $bk<0$ for odd $k$ thus,

$ϕk=−π2$

Therefore,

$ϕk={π2k even−π2k odd$

Thus, the phases corresponding to $k=1,2,3,4,5,...$ are $−π2,π2,−π2,π2,−π2,...$

Use the following MATLAB code to construct the amplitude plot.

function Code_97924_19_6P_a()

for k = 1: 8

f(k) = k;

% define the amplitude

A(k) = 2/(k*pi);

end

% plot the values thus obtained

stem(f,A,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[A(8) A(4) A(2) A(1)]);

set(gca,'YTickLabel',{'1/4\pi' '1/2\pi' '1/\pi' '2/\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

end

Execute the above code to obtain the amplitude plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 4

Interpretation: The above plot shows the amplitude plot for the sawtooth wave as shown in the figure provided.

Use the following MATLAB file can be used to construct the phase plot.

function Code_97924_19_6P_b()

for k = 1: 8

f(k) = k;

% define the phase

P(k) = (-1)^k*(pi/2);

end

% plot the values thus obtained

stem(f,P,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[-pi -pi/2 0 pi/2 pi]);

set(gca,'YTickLabel',{'-\pi' '-\pi/2' '0' '\pi/2' '\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

ylim([-pi-0.1, pi+0.1]); grid on

end

Execute the above code to obtain the plot as,

Numerical Methods for Engineers, Chapter 19, Problem 6P , additional homework tip 5

Interpretation: The above plot shows the phase line spectra for the sawtooth wave as shown in the figure provided.

Answer 13

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Construct amplitude and phase line spectra for Prob. 19.4.

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